Can somebody help me with this problem. Like asap

Andre45 [30]

Answer:

First, we are going to find the sum of their age. To do that we are going to add the age of Eli, the age Freda, and the age of Geoff:

The combined age of Eli, Freda, and Geoff is 40, so the denominator of each ratio will be 40.

Next, we are going to multiply the ratio between the age of the person and their combined age by £800:

For Eli:

For Freda:

For Geoff:

We can conclude that Eli will get £180, Freda will get £260, and Geoff will get £360.

Step-by-step explanation:

6 0

2 years ago

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Activities of tariffs

Dmitry [639]

Answer:

What??

Step-by-step explanation:

5 0

2 years ago

Ivan has 4 hours of Music on his MP3 player Jazz makes up 1/6 of the total time how many hours of jazz are on Ivan's MP3 player

Crank

So theres 4 hours and 60min equals 1 hour, so multiply 4 by 60 to get 240, so theres 240 minutes of time on the mp3. So 1/6 of that total time (240min) was on jazz, so what you do is divide 240 by 6 to get 40, so every 1/6 is 40min. So you could say he spend 40min listening to jazz, or you could say he spent 2/3 of an hour listening to jazz

5 0

2 years ago

In a recent survey about what food was your favorite take-out, it was determined that 40% of students preferred pizza, 32% of st

Nataliya [291]

Answer:

576 students choose cheeseburgers

Step-by-step explanation:

Given: In a survey, it was determined that 40% of students preferred pizza, 32% of students preferred cheeseburgers, and 28% of students preferred tacos.

To find: Number of students out of 1800 who choose cheeseburgers

Solution:

Percentage of students who preferred cheeseburgers = 32%

Total number of students = 1800

Number of students out of 1800 who choose cheeseburgers = $\frac{32}{100}(1800)=576$

8 0

2 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0

2 years ago